mm-2119 === Subject: Re: another boring critisism of Cantor's Theorem > >> Note that, assuming set theory itself is consistent, >> it is also consistent that every model of set theory >> is countable, > Then what you meant to say was simply > false. > It doesn't follow that it is consistent > with set theory that every model of set theory is countable. Are you sure about that? Set theory is blind to the size of its model. === Subject: Re: another boring critisism of Cantor's Theorem > > >> Note that, assuming set theory itself is consistent, >> it is also consistent that every model of set theory >> is countable, > > Then what you meant to say was simply > false. > > It doesn't follow that it is consistent > with set theory that every model of set theory is countable. > > Are you sure about that? Set theory is blind to the > size of its model. It's consistent with ZFC that ZFC is inconsistent, and therefore has no models at all. In which case it's vacuously true that every model is countable. But I don't think this is what you had in mind. It's not consistent with ZFC+Con(ZFC) that every model of ZFC is countable. If ZFC is consistent, it has models of arbitrarily large cardinality by the Loewenheim-Skolem theorem. === Subject: Re: another boring critisism of Cantor's Theorem David Petry says... >> It doesn't follow that it is consistent >> with set theory that every model of set theory is countable. >Are you sure about that? Set theory is blind to the >size of its model. Yes, but I think you can prove in ZFC that if a countable language has any infinite model, then it has infinite models of all cardinalities. So it's impossible for a theory to only have countable models. -- Daryl McCullough Ithaca, NY === Subject: Set-theoretic combinatorial problem Here's a simple combinatorial problem whose solution eludes me. Let S be a finite set. Define a partial subset of S as an ordered pair of sets such that: 1) x is a subset of S; 2) y is a subset of S; 3) x and y are disjoint. If S has N elements, how many partial subsets does it have? (Clearly there are more partial subsets than subsets in the usual sense; but how many more?) Mike Carroll === Subject: Re: Set-theoretic combinatorial problem Ok, I got it. That's N independent choices with 3 options each instead of 2. So S has (3**n) partial subsets. How'd I miss that? Must be gettin' old. === Subject: Re: Set-theoretic combinatorial problem >Ok, I got it. That's N independent choices with 3 options each instead >of 2. So S has (3**n) partial subsets. How'd I miss that? Must be >gettin' old. Just what I was thinking. Each element is either: 1) in X, 2) in Y or 3) in neither. ...and I thought I was going to be useful... -- BELANGER -- Be seeing you. === Subject: Re: Set-theoretic combinatorial problem 3QLpj-NoP*NzsIC,boYU]bQ]H'y<#4ga3$21: > Here's a simple combinatorial problem whose solution eludes me. > > Let S be a finite set. Define a partial subset of S as an ordered > pair of sets such that: > 1) x is a subset of S; > 2) y is a subset of S; > 3) x and y are disjoint. > > If S has N elements, how many partial subsets does it have? (Clearly > there are more partial subsets than subsets in the usual sense; but > how many more?) This looks like homework to me. Maybe you could try enumerating all partial subsets when |S|=0, |S|=1, and |S|=2, and see whether you detect a pattern in the numbers you get. Make sure you remember that the empty subset is disjoint from anything else including itself. -- David Eppstein http://www.ics.uci.edu/~eppstein/ Univ. of California, Irvine, School of Information & Computer Science === Subject: Re: =?ISO-2022-JP?B?GyRCOiNGfCRIJCQkJkZ8JHJANiQvQDUkNyQvQDgkLRsoQg==?= =?ISO-2022-JP?B?GyRCJGgkJhsoQg==?= > >.83E.83202[Eth].82å.82[Hyphen].82é 202.82.82.88.82ç.82[CapitalADoubleDot ].82í.82ô.82©.82.81 202á.81B > >>joke .82ê.83E.83202í.88.90.8e.922 02é.82í.82.81.82á.82 ».81A.89.bd.82©.81H >> >.83P.83.93.83u.83.8a.83b.83W.8du.89.89.82ê joke .82é.82.96.82.81.82á.82[Cop yright].81H.81@(.81K.84t.81K) >Your Cambridge University lecture Is it joke? :-) > You are a Philipino, aren't you? This is known by too-correct English. Because translating machine gives us TOO-CORRECT English. -- v(^-^)v pupupu! === Subject: Re: Lenstra's ECM >Could somebody explain to me how Lenstra's ECM of factorisation works >or point me towards a place where I could find out more about it. DMcG. > > See P. Montgomery's paper > Speeding the Pollard & Elliptic Curve Methods of Factorization > Math. COmp. Jan, 1987 > > Very quickly: Consider a random elliptic curve over GF(p). > By Hasse's theorem the order of the variety is bounded in > [p - 2sqrt[p] + 1, p + 2sqrt[p] + 1]. A Theorem of Deuring says > that that all orders in this interval are achieved and that the > distribution of orders follows a sin^2 distribution. Bob, is the sin^2 distribution really in Deuring. I had thought that this was a theorem of Bryan Birch in the 1960's. If I'm wrong, I'd definitely appreciate a reference so that I can attribute it properly in the future. (Of course, for the reductions mod p of a fixed non-CM elliptic curve over Q, it's the Sato-Tate conjecture, but that's a different question and came much later.) Joe === Subject: Re: Illegal to do research on cryptography? > In a nutshell, my system creates a feeling of security, particularly since > the encrypted number changes each time for the same credit card. It has > protected me and my customers against all real attacks so far. so you intend to make a business out of creating a false sense of security... this can be summed up in 2 words - snake oil... -- we're the first ones to starve, we're the first ones to die the first ones in line for that pie in the sky and we're always the last when the cream is shared out for the worker is working when the fat cat's about === Subject: Re: Illegal to do research on cryptography? >so you intend to make a business out of creating a false sense of >security... >this can be summed up in 2 words - snake oil... Snake oil, in the form of misleadingly-advertised SSL certificates, a multimillion-dollar business. -- Rahul === Subject: Re: Illegal to do research on cryptography? > >>so you intend to make a business out of creating a false sense of >>security... > >>this can be summed up in 2 words - snake oil... > > Snake oil, in the form of misleadingly-advertised SSL certificates, > a multimillion-dollar business. it's not clear, since you snipped the context of what i was replying to, but your response is a red herring... -- we're the first ones to starve, we're the first ones to die the first ones in line for that pie in the sky and we're always the last when the cream is shared out for the worker is working when the fat cat's about === Subject: Re: Illegal to do research on cryptography? > >>so you intend to make a business out of creating a false sense of >>security... > >>this can be summed up in 2 words - snake oil... > > Snake oil, in the form of misleadingly-advertised SSL certificates, > a multimillion-dollar business. Another side issue to the thread. I too agree that the PKI *industry* is a huge scam. That doesn't mean the technology is flawed though. Hell I wouldn't have much problems with using a site that didn't have a CA signed public key. Tom === Subject: Re: Illegal to do research on cryptography? > I am open to your suggestion. I just don't have much time to go through the > literature. Another solution consists of selling your algo if you have good > stuff. But how would I know that you have good stuff, unless I spend many > hours doing research, when I do not have much time? > > I used to be very good mathematically speaking and very bad from a business > viewpoint. Now I'm not as good mathematically as I used to be (I'm not bad > either), but I've became a good salesperson. So obviously I want to exploit > my skills. And one of the skills of a good salesman is to carefully listen > to what people say. with crypto it makes good business sense to use algorithms that have been agreed upon as standards after careful examination by official bodies - it's quite a bit easier to sell people on the security of such an algorithm than on one you made in your spare time... you say you've become a good salesperson, this should make sense to you... descriptions of the algorithm and even sample implementations in very short order... -- we're the first ones to starve, we're the first ones to die the first ones in line for that pie in the sky and we're always the last when the cream is shared out for the worker is working when the fat cat's about === Subject: Re: Illegal to do research on cryptography? > Any of these will give you an introductory look into cryptography. And > more importantly show you how easy it is to get crypto wrong, e.g. every > (?) cipher used during world war II has been broken. Countries trusted > their most important secrest to these ciphers expecting them to be > secure. One-time pads are still secure. Not very efective for high-volume traffic, however. === Subject: Re: Question about a sequence equation > What is the equation for this sequence 1 2 5 10 21 42 85, etc. > Could you explain why it is? Looks like each term is twice the previous, plus c, where c alternates between zero and one. c could be written as half of (one plus ((minus one) to the n)). Then there are standard techniques for solving such recurrence relations - any good text on discrete mathematics should help. -- Gerry Myerson (gerry@maths.mq.edi.ai) (i -> u for email) === Subject: Re: Question about a sequence equation >> What is the equation for this sequence 1 2 5 10 21 42 85, etc. >> Could you explain why it is? >Looks like each term is twice the previous, plus c, >where c alternates between zero and one. >c could be written as half of (one plus ((minus one) to the n)). >Then there are standard techniques for solving such recurrence >relations - any good text on discrete mathematics should help. >-- >Gerry Myerson (gerry@maths.mq.edi.ai) (i -> u for email) Following Gerry's observation, write the sequence in binary (base 2). -- John Adams served two terms as Vice President and one as President, but lost reelection. Later his son became President despite losing the popular vote. That son lost his reelection attempt badly. Now history is repeating itself. pmontgom@cwi.nl Microsoft Research and CWI Home: San Rafael, California === Subject: Re: Question about a sequence equation 3QLpj-NoP*NzsIC,boYU]bQ]H'y<#4ga3$21: > > What is the equation for this sequence 1 2 5 10 21 42 85, etc. > Could you explain why it is? > > Looks like each term is twice the previous, plus c, > where c alternates between zero and one. > > c could be written as half of (one plus ((minus one) to the n)). > > Then there are standard techniques for solving such recurrence > relations - any good text on discrete mathematics should help. The same sequence is generated by other recurrences (with appropriate base cases): s_i = 2^i + s_{i-2} s_i = ceiling((i+1)/2) + sum_{jy<#4ga3$21: > What is the equation for this sequence 1 2 5 10 21 42 85, etc. > Looks like each term is twice the previous, plus c, > where c alternates between zero and one. > > c could be written as half of (one plus ((minus one) to the n)). > > Then there are standard techniques for solving such recurrence > relations - any good text on discrete mathematics should help. > > The same sequence is generated by other recurrences > (with appropriate base cases): > s_i = 2^i + s_{i-2} > s_i = ceiling((i+1)/2) + sum_{j looks like a reasonable place to start. It looks like the method in section 5 of that page applies directly to this last recurrence. -- David Eppstein http://www.ics.uci.edu/~eppstein/ Univ. of California, Irvine, School of Information & Computer Science === Subject: Re: trisecting an angle with a straightedge(not marked)&compass alone > but I have found out a seven step method > by which an angle(obtuse or acute) can be trisected with a compass and straightedge( not marked)alone. No, you didn't. Doug === Subject: Re: why are mathematicians interested in fluid dynamics? > I thought this kind of thing was left to the engineers? Hey, mathematicians like their beer as much as anyone does. -- Gerry Myerson (gerry@maths.mq.edi.ai) (i -> u for email) === Subject: Re: sums of unit fractions > Nice. There may be a constructive proof in Johannessen and Sohus, > On unit fractions, Nordisk Mat. Tidskr. 22 (1974) 103-107, 135, > MR 55 #252. The review in Math Reviews says that for all integers > p and k, p / (k p! + 1) requires p unit fractions. The paper is in > Norwegian. I must have misunderstood the review, as what I have written is nonsense. E.g., 4/25 = 1/10 + 1/25 + 1/50 shows that it's false for p = 4, k = 1. In fact, if the Erdos-Straus conjecture is correct (it states that for every n there exist x, y, z such that 4/n = 1/x + 1/y + 1/z) then p / (k p! + 1) can't require p unit fractions for any p > 3. -- Gerry Myerson (gerry@maths.mq.edi.ai) (i -> u for email) === Subject: principal ideal in Z_p[x] Prove that I={ fin Z_p[x] : f(x)=0 for all xin Z_p} is precisely the principal ideal J=(x^p-x). I am able to show Jsubset I easily since f(x)in J implies f(x)=(x^p-x)g(x) for some g(x)in Z_p[x], and obviously x^p-x=0 for all xin Z_p. Therefore f(x)in I. Inclusion the other way seems very easy, but I think I may have gone about it the wrong way. If f(x)in I, then by the division algorithm f(x)=x(x-1)...(x-(p-1))g(x) for some g(x)in Z_p[x]. At this point I see that x(x-1)...(x-(p-1)) agrees pointwise with x^p-x, so can I simply say that x(x-1)...(x-(p-1))=x^p-x, and therefore f(x) is of the form (x^p-x)g(x), hence in the ideal J? === Subject: Re: principal ideal in Z_p[x] > > Prove that I={ fin Z_p[x] : f(x)=0 for all xin Z_p} is precisely the > principal ideal J=(x^p-x). > > I am able to show Jsubset I easily since f(x)in J implies > f(x)=(x^p-x)g(x) for some g(x)in Z_p[x], and obviously x^p-x=0 for > all xin Z_p. Therefore f(x)in I. > > Inclusion the other way seems very easy, but I think I may have gone > about it the wrong way. If f(x)in I, then by the division algorithm > f(x)=x(x-1)...(x-(p-1))g(x) for some g(x)in Z_p[x]. At this point I > see that x(x-1)...(x-(p-1)) agrees pointwise with x^p-x, so can I > simply say that x(x-1)...(x-(p-1))=x^p-x, and therefore f(x) is of the > form (x^p-x)g(x), hence in the ideal J? Below are slightly simpler (and more general) approaches. Hint: I = (f) > (x^p-x) properly => f | x^p-x properly Alternatively, if you don't already know Z_p[x] is a PID: If I > J, choose f in I, not in J. Consider f mod x^p-x It is a nonzero element of I of degree < p so .... Neither proof requires x^p - x = x(x-1)(x-2)...(x-(p-1)) only that a polynomial over a field (or over a domain) can't have more roots than its degree (not true if there are zero divisors: ab=0 => ax = 0 has > 1 root: x = 0,b e.g. over integers mod 4, 2x = 0 has > 1 root: x = 0,2). -Bill Dubuque === Subject: Re: principal ideal in Z_p[x] >Prove that I={ fin Z_p[x] : f(x)=0 for all xin Z_p} is precisely the >principal ideal J=(x^p-x). >I am able to show Jsubset I easily since f(x)in J implies >f(x)=(x^p-x)g(x) for some g(x)in Z_p[x], and obviously x^p-x=0 for >all xin Z_p. Therefore f(x)in I. >Inclusion the other way seems very easy, but I think I may have gone >about it the wrong way. If f(x)in I, then by the division algorithm >f(x)=x(x-1)...(x-(p-1))g(x) for some g(x)in Z_p[x]. At this point I >see that x(x-1)...(x-(p-1)) agrees pointwise with x^p-x, so can I >simply say that x(x-1)...(x-(p-1))=x^p-x, and therefore f(x) is of the >form (x^p-x)g(x), hence in the ideal J? Since f(x) has degree p and x(x-1)...(x-(p-1)) has degree p, then x^p-x = x(x-1)...(x-(p-1))h(x), where h(x) is a nonzero polynomial of degree 0. It follows that h(x) is a nonzero constant, and so x(x-1)...(x-(p-1)) is a multiple of x^p-x, and so is an element of J. Further, by taking the coefficient of x^p on both sides, h(x) = 1, so that x^p-x = x(x-1)...(x-(p-1)). David But I'm always true to you, darlin', in my fashion, Yes, I'm always true to you, darlin', in my way. -- Lois Lane ----- === Subject: Re: principal ideal in Z_p[x] >>Prove that I={ fin Z_p[x] : f(x)=0 for all xin Z_p} is precisely the >>principal ideal J=(x^p-x). >>I am able to show Jsubset I easily since f(x)in J implies >>f(x)=(x^p-x)g(x) for some g(x)in Z_p[x], and obviously x^p-x=0 for >>all xin Z_p. Therefore f(x)in I. >>Inclusion the other way seems very easy, but I think I may have gone >>about it the wrong way. If f(x)in I, then by the division algorithm >>f(x)=x(x-1)...(x-(p-1))g(x) for some g(x)in Z_p[x]. At this point I >>see that x(x-1)...(x-(p-1)) agrees pointwise with x^p-x, so can I >>simply say that x(x-1)...(x-(p-1))=x^p-x, and therefore f(x) is of the >>form (x^p-x)g(x), hence in the ideal J? >Since f(x) has degree p This part should have read Since x^p-x has degee p. >and x(x-1)...(x-(p-1)) has degree p, then >x^p-x = x(x-1)...(x-(p-1))h(x), where h(x) is a nonzero polynomial of >degree 0. It follows that h(x) is a nonzero constant, and so >x(x-1)...(x-(p-1)) is a multiple of x^p-x, and so is an element of J. >Further, by taking the coefficient of x^p on both sides, h(x) = 1, so >that x^p-x = x(x-1)...(x-(p-1)). David But I'm always true to you, darlin', in my fashion, Yes, I'm always true to you, darlin', in my way. -- Lois Lane ----- === Subject: Re: How to hire a mathematician? > I wonder how much time (percentage) the average research mathematician > wastes? i.e., How would you commission (contract) a good math guy to > do a math research project for a certain $? Let's assume a decent wage > without wasted time is $X/hr. If the ave. wasted time were Y%, should > we justly pay him only $XY/100? Should we add a non-timely completion > penalty clause, etc.? > How do we determine a fair commission (wage)? vs payoff? or? Not like > Lawyers I hope. Flat contract fee with explicit expectation of productivity > (preferably with bonuses on completions) and termination conditions, > plus royalties. Put the near-term money against timely achievement. > Do not go cheap if he cracks the nut early - or he'll take his sweet > time next time. If he makes you $100 million and you don't toss back > some *serious* folding green, you are an ass. He'll visit your > competitors. Don't bind the mouths of the kine who tread the grain. > They will work against you. Remember that the future is after your > ass, always. How do you value a fellow who sits around for five years apparently > doing nothing, then solves a classic problem nobody else can? > Efficient management of reseach safely and timely results in nothing. > An easy problem will be cracked by a mediocre mind and a PERT chart. > If you are wandering into the unknown, he doesn't know how to get > there and you don't know how to tell him to get there. Write off the > money and go for it, or chicken out. If you want to find the bottleneck, the first place to look is at the > top of the bottle. How can Management be held responsible when they > never lift the heavy end? Management makes decisions, workers make > mistakes. Why is Google an unassailable powerhouse, one of the few dotcoms to > make nine figures? It hires nerds based on academic and standardized > test grades. It allows then to screw around a lot. If you want your > resident geniuses to do great things, get them bored. The best of > them cannot tolerate the silence. Pay them or fire them. If you > start playing piddly MBA games with your brains you will be crushed by > an unsahven guy tinkering in his garage - or an ex-employee with a new > job. Hire first rate people. Toss money and/or perqs (ask what they want) > at them when and as they succeed. Go into your office, close the > door, sit at your desk, and commence to sweat blood. Imagine how > surprised Emperor Hirohito was at Hiroshima and Nagasaki, ditto > Saddamn Hussein and stealth. > > I am somewhat surprised by your answers. I expected a more > sophisticated, more mathematical, more elegant solution proposal. > Something more arcane, perhaps involving game theory, O.R. or the > like. But your approach seems nore a practical, gut-feel style; still, > one that I appreciate. However, I prefer a little more risk management > in my investments. I hate sweating blood and sleepless nights. > Speaking of the atom bomb project, who decided to hire Oppenhiemer, > his boss, et al? They certainly got the job done well! But, I'm not > sure the world is better off because of their accomplishments and > success with that program. I guess we'll never know what would have > been the best overall solution. Honestly, my gut instinct is that we > humans should never have screwed around with nuclear radiation; not in > this world or time-frame! > > If you want the best (do bother to define that) possible solution to > hiring a mathematician, hire a mathematician to derive it. You sure > don't want to hire an MBA. All you will obtain in the latter case is > the best possible resume for the MBA appyling for his next > appointment. All female and minority hires who are not prodigies are > rejected flat out. Diversity is inimical to net retained earnings. > > If you need labor, you manage. If you need genius, you wing it. > Every advance will leave a fraction of incompetent humanity behind an > increment. They will lose. Tough nuggies. Evolution is a blast as > long as you are one of the survivors. > > The Watts slum in Los Angeles, CA is notorious for its riots and > lowest class inhabitants; ditto Brazilian favelas. Have you ever been > to Watts? It is individual bungalos on fenced small plots of land. > Every resident has hot and cold potable running water, electricity, > indoor plumbing, TV, radio... They are well-fed, well-clothed, have > ample medical care, free public education through grade 12, free > public libraries. They live like compared to their betters. > > Brazilian favelas are swamps of mud, human excrement, lean-tos and > decaying buuildings. Water might be one dubious faucet for 500 > people. They are starved and sick, without education. They wear > rags. > > That is the difference between a slum in a technological society and a > slum in the Third World. That is the difference between the dregs of > mathematics, science, technology, engineering... and trusting in > priests. We grab everything we can, we advance as far and as fast as > possible in everything. We care about decimal places. Our mistakes > and garbage middens are as nothing compared to the alternative. > > Three Mile Island vented a few curies of volatiles, mostly iodine and > xenon. Chernobyl vented a few billion curies of the entire Periodic > Table. Neither situation was a happy thing. If you cannot quantitate > the difference, you are a Liberal. > > If it isn't mathematics it is merely opinion. The universe doesn't > care about your opinions. Does the universe care at all, about anything? It might be nice if something out there really cared! They say Jesus, the Christ, is out there preparing fancy places for His faithful followers. Are you one the them? Brother AL? === Subject: Re: How to hire a mathematician? > I wonder how much time (percentage) the average research mathematician > wastes? i.e., How would you commission (contract) a good math guy to > do a math research project for a certain $? Let's assume a decent wage > without wasted time is $X/hr. If the ave. wasted time were Y%, should > we justly pay him only $XY/100? Should we add a non-timely completion > penalty clause, etc.? > How do we determine a fair commission (wage)? vs payoff? or? Not like > Lawyers I hope. > > OK, I know you're trolling but... How can you tell what wasted time is? Nope, I was Not trolling. So, perhaps you aren't always 100% reliable in your judgement? Fair question though! ..what wasted time is? I don't know.vThat's part of my problem, so subjective! In my own experience, I think I wasted about half of my life! Based on my GPA (4.0) in advanced math, I thought I was special at it. But, I worked on FLT in fits and starts for over 20 yrs, and never got past exp n=2 with any success :-( However, by failing that, I did learn my limitations to some degree. AYE, 'tis a Bitter pill to swallow; to discover that you're probably mediocre at your favorite pastime or discipline - after all. === Subject: Re: Erdos Number Revised By sci.math... > It occurred to me... > > ;) > > If 2 mathematicians who have both posted to the same sci.math thread > can be considered to have coauthored a paper, then the Erdos numbers > of many mathematicians might be lowered. > > I myself, for instance, might have a finite Erdos number with this > sci.math-standard. > > Which posters here on sci.math have a resonably low Erdos number, if > any? > > Leroy Quet This would be rather stupid. Trolls with no mathematical education at all would soon have 2s. Don't you know there are some yet-unpublished papers of which Erdos is a co-author? If one of their stewards posted one of them here as their way of publishing them, then everyone could hop aboard and all of usenet could have 1s. On a more serious note, your post inspired me to analyze the Erdos numbers, which are (you realize) dynamic and ever-changing, and I managed (to my great delight) to prove that as the population of earth approaches infinity, the ratio of people with countably infinite Erdos numbers to people with uncountable Erdos numbers would approach 1/5pi. In proving this I had to rely on some unpublished letters which Erdos and I exchanged, and so I will be listing him as coauthor when I publish this remarkable result. Your dear friend, Nathan the Great Age 11 :-) === Subject: Re: Erdos Number Revised By sci.math... > > It occurred to me... ;) If 2 mathematicians who have both posted to the same sci.math thread > can be considered to have coauthored a paper, then the Erdos numbers > of many mathematicians might be lowered. I myself, for instance, might have a finite Erdos number with this > sci.math-standard. Which posters here on sci.math have a resonably low Erdos number, if > any? Leroy Quet > > Now you've done it. James Harris now has an Erdos number. Sure, and James Always in error, never in doubt! Harris has a slowly executing Java program to prove it is an irrational recurring non-cardinal integer. -- Uncle Al http://www.mazepath.com/uncleal/qz.pdf http://www.mazepath.com/uncleal/eotvos.htm (Do something naughty to physics) === Subject: Re: Erdos Number Revised By sci.math... > Other than Ron Graham & Fay Chung are there any other married couples > where both have an Erdos number of 1? Possibly George & Esther Szekeres. -- Gerry Myerson (gerry@maths.mq.edi.ai) (i -> u for email) === Subject: Line tangental to the surface of the earth. Earlier I asked the question: How would one find a line tangental to the earth's surface if you had the location of a point on the earths surface in lat, lon and elevation? I would like to add, that I would like to be able to have the point on the surface of the earth go through the center of the earth, The line that I am looking for would be a vector normal to the vector that goes through the center of the earth. How would one do this? -- Sean === Subject: Are Modern Astronomers Low-Grade Morons? Comments: This message probably did not originate from the above address. It was automatically remailed by one or more anonymous mail services. You should NEVER trust ANY address on Usenet ANYWAYS: use PGP !!! Get information about complaints from the URL below unreachable: It is because it has been flooded by , please contact -----BEGIN PGP SIGNED MESSAGE----- Sadly, the answer is Yes. Modern astronomers are ignorant, and they prefer to stay that way. Only in death will they finally admit to their servile cowardice and incompetence (assuming their IQ > 70, which at this point is doubtful). I'd offered substantial reward monies for going on a year now, but not one astronomer was willing or able to accept my generous cash offer. They claim to be intelligent, but their deafening silence proves that they are lame-brained and uneducated. As a result, I've withdrawn the reward $$ I'd offered, since modern astronomers are plain worthless. *Min's $0.00 (zero cents) Reward For Orthodox Astronomers: Very Truly Yours, Daniel Joseph Min -----BEGIN PGP SIGNATURE----- iQA/AwUBQHIPQpljD7YrHM/nEQKkVACfVTy/OmmUxvTOFKoJ/PKGmxSHAugAnjaJ UidPHydCAwVTGk51mghYg6O7 =l3wo -----END PGP SIGNATURE----- === Subject: Re: Are Modern Astrologers Low-Grade Morons? Comments: This message probably did not originate from the above address. It was automatically remailed by one or more anonymous mail services. You should NEVER trust ANY address on Usenet ANYWAYS: use PGP !!! Get information about complaints from the URL below unreachable: It is because it has been flooded by , please contact > -----BEGIN PGP SIGNED MESSAGE----- > Sadly, the answer is Yes. Modern astrologers are ignorant, > and they prefer to stay that way. Only in death will they > finally admit to their servile cowardice and incompetence > (assuming their IQ > 70, which at this point is doubtful). So says our resident morphing bumwipe. === Subject: Re: Are Modern Astronomers Low-Grade Morons? A-B: C Disapproved: Good Housekeeping > -----BEGIN PGP SIGNED MESSAGE----- > > Sadly, the answer is Yes. Not all of them, Dannie, just those like you. === Subject: Re: Hille's lower bound for all primes of ~ 1.72864 (ordered factorisation) > Let F(n) be the number of ordered factorizations of n. > The lower bound says that for any exponent b<1.72864, > we have lim sup F(n)/n^b = infinity. > Put another way, for any b<1.72864, for any C>0, > there is an n for which F(n) > c n^b. > > The limit point d=1.72864 is zeta^{-1}(2), that is, the solution of > sum_n 1/n^d = 2. > > > Moshe Lewenstein and I have recently given a fairly straightforward > constructive proof, which is still in draft form; I can send you > a copy in a couple of weeks when we have finalized it. > Remind me in a few weeks. Hi Don I would be very interested in this too. I will make a post following this with the subject:Ordered Factorization, Integer Sequence 074206 and Harmonics of the Universe. Please read that and give me your comments. I am not a mathematician really, but have been muddling my way through this for some years. Ray Tomes === Subject: complex analysis problem. hello.......dear sir~ show that if c is a complex number such that |c|>e , then the equation c(z^n) = e^z has n roots, counting multiplicities, inside the circle |z|=1 ---------------------- um....i know the Rouch theorem. thus suppose that f(z) = c(z^n) g(z) = -e^z thus, f,g is analytic |z|<=1 and |f(z)| > |g(z)| on |z|=1 thus, f(z) and f(z)+g(z) have the same number of zeros inside |z|<1 but, i don't know that f(z) = c(z^n) has n roots on |z|<1 thus, i can't finish that problem. advice...please...thank you. === Subject: Re: complex analysis problem. >hello.......dear sir~ >show that if c is a complex number such that |c|>e , >then the equation c(z^n) = e^z has n roots, >counting multiplicities, inside the circle |z|=1 >---------------------- >um....i know the Rouch theorem. >thus >suppose that >f(z) = c(z^n) >g(z) = -e^z >thus, f,g is analytic |z|<=1 and |f(z)| > |g(z)| on |z|=1 >thus, f(z) and f(z)+g(z) have the same number of zeros inside |z|<1 >but, i don't know that f(z) = c(z^n) has n roots on |z|<1 f(z) has an n-fold zero at 0, i.e. 0 is a root of f(z) = 0, n times over. David But I'm always true to you, darlin', in my fashion, Yes, I'm always true to you, darlin', in my way. -- Lois Lane ----- === Subject: Ordered Factorization, Integer Sequence 074206 and Harmonics of the Universe The following is an outline of what is on the web page listed below and concerns the interesting maths of Number of Ordered Factorisations of numbers and the application to the structure of the universe which results because Maxwell's equations under GR are non-linear and therefore all wave structures in the universe must develop harmonics. After decades on this project I can state that this method bears fruit in explaining the structure of the universe and why there are the various structures that we observe at the various scales with astronomical ratios between them. All this from pure maths with no arbitrary physical constants. Each integer h can be factorised in C(h) ways where C(1)=1 and for larger h, C(h) is the sum of C(i) where i is a factor of h excluding h itself. This leads to the series: 0,1,1,1,2,1,3,1,4,2,3,1,8,1,3,3,8,1,8,1,8,3,3,1,20,2,3, ... For 15 years I have been working with various aspects of the maths and consequences of how many ways a number may be factorised. Then my friend Mountain Man found that I was not the only one doing this and sent me links to the sites: http://www.research.att.com/cgi-bin/access.cgi/as/njas/sequences/eisA.cgi?An um=A074206 http://mathworld.wolfram.com/OrderedFactorization.html where the same thing is called Ordered Factorization and Integer Sequence 074206. OK, it mightn't sound complicated enough for spending 15 years on so here is the reason, it is to do with the applications of the result and my need to go to very large numbers in the calculations. I have proposed the Harmonics Theory which states: The universe is a standing wave which develops harmonics and each of these is also a standing wave that does the same. This means that each harmonic h can occur in C(h) ways where C(1)=1 and C(h) is the sum of C(i) where i is a factor of h excluding h itself. There is another aspect of harmonics and that is that higher order harmonics tend to be weaker (due to non-linearity producing them that way) and if they are assumed to follow an inverse power law like h^-k with k a constant, then the power in the h-th harmonic is C(h)*h^-k because h^-k comes out the same for all paths through the harmonics. The harmonics theory is interested in all the harmonics, h, which have extremely high values for C(h) for the size of h because they will dominate the energy of the universe at that scale. I have called the very strongest harmonics that are linked by prime ratios the main-line and it goes 1, 2, 4, 12, 24, 48, 144, 288, 1440, 2880, 5760, 17280, 34560, .... where the ratios of adjacent terms are 2, 2, 3, 2, 2, 3, 2, 5, 2, 2, 3, 2, 2, ... and although there are some cases where my very strongest harmonics that are linked by prime ratios may not be exactly a mathematically precise one, the idea bears fruit for my purposes. This last series has some very interesting properties: It seems to be the result of determining all the multiples of p*ln(p) for each prime and then selecting the next lowest unused one to go into the list (I hope that makes sense - it means that each prime appears with frequency which is the inverse of p*ln(p) in the list). Actually it gradually departs from this relationship at higher h. By this means the exact calculation which I have carried out to beyond 10^50 for relatively strong harmonics can be carried on approximately to much larger numbers. I am interested in being able to the exact calculation to larger numbers such as up to 10^80. The result gives a very good fit to the energy distribution of the universe at all scales with each type of observed structure corresponding to especially strong harmonics such as 34560 (2^8*3^3*5) and others near powers of that number (but with more other primes in their factorisations). These numbers when interpreted as harmonics show the observed *distance* structures of: Hubble scale, galaxies, stars, planets, moons, ... cells, atoms, nucleons. Using the theory I have made many successful predictions ranging from geological cycle periods and galaxy redshift periodicities through to discovery. See my pages at: http://homepages.ihug.co.nz/~ray.tomes/story.htm The especially strong harmonics at ratios near powers of 34560 may not be at all obvious in its meaning. To make it clearer I need to introduce a concept that I call Ideal harmonics. An Ideal harmonic is one that has in its factorisation prime indices which when 1/2 is added to each one are in inverse proportion to p*ln(p). An example is 34560 which is 2^8*3^3*5^1 and so the indices plus 1/2 are 8.5, 3.5, 1.5 respectively. The values of p*ln(p) for the primes 2, 3, 5 are about 1.386, 3.296, 8.047 and multiplying these by the indices plus 1/2 gives 11.78, 11.54, 12.07 respectively which are nearly constant. Therefore 34560 is an almost ideal harmonic, whereas one like 1683 would be very far from ideal. Actually the higher h gradually depart from this p*ln(p) relationship, but the concept of ideal harmonoics being ones with the very best combination of prime indices can still be used . When the size of C(h) are compared for different numbers h of the same order of magnitude as each other then it can easily be seen that this ideal harmonics concept works. It is then not too big a stretch to apply it to the whole sweep of harmonics and that is when it is found that 34560 and many other numbers near powers of that number come close to being ideal. I hope that you find this interesting and that some mathematicians can be encouraged to work on the questions of methods of computing C(h) for very large h and also making proofs of the accuracy of the ideal harmonics concept which has so far only been developed by inspection and cunning correlations. I have not yet finished work on this page, but as Mountain Man has raised the subject of NOF (Number of Ordered Factorisations) it seems time to release this and get some comments. Please see http://homepages.ihug.co.nz/~ray.tomes/maths.html Ray Tomes === Subject: String Theory (analogous to Number Theory) Number Theory is amazing in its problems of every degree of complexity and difficulty, and the frequent lack of correlation between complexity and difficulty. But to a computer programmer, Number Theory is just programs that use only functions + - * / ^ and branching (control) mechanisms. For example, FLT is just a program that searches for a solution to A^N+B^N=C^N where N>2, and trying to determine if it will ever halt. But computer programmers also like to play with non-numeric string functions: concatenate two strings, determine if one string is contained in another or its leftmost occurrence, breaking a string apart at certain characters, etc. My question is: Can we construct a String Theory that is analogous to Number Theory? Then we would have string equations using the aforementioned string functions, questions about the existence of a string that meets a given predicate, searches for efficient algorithms to perform certain string functions, etc. - analogues for each of the processes that occur in Number Theory. Charlie Volkstorf Cambridge, MA === Subject: Strings and Loops & Zero Point > Note #13 > > Basic concepts in the gravitational physics of the quantum vacuum not included in the conceptions of NASA BPP or Puthoff & Co in their quest for metric engineering. > > What is the non-exotic or ordinary vacuum at a given scale of coarse-graining? > > > The non-exotic vacuum has no gravitational properties at all! This is the common sense idea of the vacuum. However, what is not understood, by even mainstream physicists today, is that this means that the non-exotic vacuum is optimally macro-quantum coherent with no trace of random micro-quantum energy density at all from any quantum field on the smooth curved space-time of Einstein's theory of gravity. Indeed, this makes the solution of the cosmological constant problem in the large-scale structure of the Universe essentially trivial. What is missing from the conceptualizations of All The King's Men is the role of macro-quantum more is different coherence of the physical vacuum. You can think of the physical vacuum in terms of the old Tiza two fluid model with the jagged edge of incoherent random micro-quantum zero point vacuum fluctuations as the normal fluid on the smooth coherent non-random superfluid. The total density is fixed by the loop quantum of area th at is dual to the string theory tension where > > Loop quantum of area = hc(String theory tension)^-1 > > Note that string theory tension has dimensions Energy/Length > > (Quantum of Area)(String Tension) ~ hc > > hc has the physical dimensions of (electric charge)^2, which is the same as (Gravity coupling)(mass)^2 > > The simple two-fluid formula > > total density = superfluid density + normal fluid density > > is equivalent to > > (Quantum of Area)^-3/2 = |Vacuum Coherence|^2 + /zpf(Quantum of Area)^-1/2 > > Einstein's local principle of equivalence of gravity as acceleration + Heisenberg's micro-quantum uncertainty principle imply that the equation for the direct gravity or anti-gravity influence of zero point energy density is the exotic vacuum local field equation > > Guv + /zpfguv = 0 > > In the weak tidal curvature field quasi-Newtonian limit, Einstein's exotic vacuum local field equation is the Poisson equation > > > The RHS source term is a special case of (G*/c^2)(exotic vacuum local energy density)(1 + 3w) > > where > > w = 0 for ordinary matter in the Hubble flow > > w = +1/3 for real photons forming far-field radiation like the CMB, or like starlight. > > w = -1 for unbalanced zero point energy density. > > This choice gives c^2/zpf above > > -1/3 > w > -1 for a quintessent field > > -1 > w for the phantom energy field which allows the Big Rip that can destroy the whole Universe in a Doomsday WMD of Apocalypse Now. > > w = -1 puts us right on the cutting edge of the possibility of total annihilation within the power of some mad man like Osama Bin Laden if he had the physics smarts. See Sir Martin Rees's Our Final Hour for those gruesome details that Andrei Sakharov worried about 20 + years ago. > > The NASA SNAP probe will measure the w of the anti-gravity cosmic dark energy field to better precision. Right now we know from Type 1a supernovae that w = -1 to 2% error. > > The Blackett effect is the conjecture that for the real on-mass-shell lepto-quarks: > > (electric charge)^2 = (Zero point energy induced strong-short range gravity coupling)(rest mass)^2 > > However, at the level of Gm^2 gravity as a quantum theory is not renormalizable. > > It is renormalizable at the level of a gauge theory. A similar thing happens in the weak force. However, the gauge bosons are not spin 2 gravitons but are spin 1 quanta from locally gauging the translation subgroup of the conformal group. They correspond to quanta of string length that are dual to the quantum of area. This duality gives the world hologram idea of t'Hooft and Susskind that generalizes the Bekenstein-Hawking black hole thermodynamics. Essentially, the independent geometrodynamic degrees of freedom of a 3D slice of space-time are 2D. Take a chunk of 3D hypersurface of 3D volume V with 2D boundary area A. The number of geometrodynamic degrees of freedom or BITS is not V/(Planck volume) but is A/Quantum of Area). > > Consider the Universe we can see with our telescopes. We are looking at the cosmic past light cone. We can see back, in principle, to a distance of the order of c/Ho where Ho is now measured accurately to 2% to correspond to Ho^-1 = 13.7 billion years. This gives c/Ho ~ 10^28 cm. > The area A of the volume V that we can see in the retarded signals from the past is therefore of the order of 10^56 cm^2. The thermodynamic entropy S of the entire Universe in this past light cone 3D chunk (it is not a space-like chunk) is > > S = kB (c/H)^2/(Quantum of Area) > > Ho is H at this moment in the cosmic expansion that is accelerating from the anti-gravity field of dark energy. This explains the irreversible arrow of time of the Second Law of Thermodynamics in terms of the expansion of the the 3D space of the Universe's past light cone because H is a function of the global cosmic time t that can be parametrized by the cosmic black body radiation temperature T whose correlation function spherical harmonic power spectrum over the celestial sphere (sky) is measured with the NASA WMAP space probe. > > So the ordinary non-exotic vacuum obeys > > /zpf = 0 > > Its vacuum coherence is optimal at > > (Quantum of area)|Vacuum Coherence|^2 = 1 > > sitting in the metastable bottom of the well of the Mexican Hat potential for the covariant Landau-Ginburg field equation for the vacuum coherence field. You find none of these ideas in Hal Puthoff's PV theory, which is a puny 100 lb weakling construction (compared to this Charles Atlas model :-)) based on the incomplete analogy of the metric with the path of light through an inhomogeneous dielectric. Hal is not even able to solve the problem of the gravity field of a rotating mass and he has no explicit description of zero point energy density in his PV model which has no black hole event horizons and no way to metric engineer with small applied electromagnetic energy densities. > > > Therefore, the basic way to metric engineer without needing huge electromagnetic energy densities is to start with the /zpf = 0 non-exotic vacuum. Have it overlap with a real superconductor. The induced /zpf away from zero is then the Josephson effect > > /zpf(induced by the real superconductor) ~ (Quantum of Area)^-1[|Vacuum Coherence||Superconductor Coherence|cosine(relative phase difference between vacuum and superconductor) > > Rotating the superconductor will induce such a phase difference as a kind of Berry phase. The rotation of the superconductor requires a Kerr metric with goi =/= 0 and although this is very weak it may have significant phase interferometry effects. This is simply a guess at this time. Hal's PV model so far is not able to describe the equivalent of goi =/= 0, which is the key field in Ray Chiao's gravity radio that is a clue. > > If one makes a Josephson interferometer enclosing area A, perhaps rotating like in the Sagnac effect > > One gets for example cosine(Flux through A/(Quantum of Flux)) > > This flux need not be only dynamical gauge force flux, but as in Foucault's pendulum described by Sir Michael Berry it can be a topological phase effect from cycling nondynamical anholonomic control parameters, which is what you get with rotation from the non-commutativity of Lorentz boosts pointing in different directions, from frame-drag (Lense-Thirring). > > Ray Chiao's gravity radio is a Lense-Thirring anholonomic frame drag effect in the far field of both real photons and real gravitons. > > We want instead, for metric engineering, to couple virtual photons in macro-quantum states (induction near fields from coils and capacitors) to the near gravity fields (not so much virtual spin 2 gravitons as their spin 1 square root gauge bosons to make a renormalizable quantum gravity i.e. spin 2 virtual gravitons as an EPR triplet pair of spin 1 gauge force bosons of Kleinert's world crystal lattice distortion field ~ (Quantum of area)^1/2 string link BITs, > > > Note #12 > > > Nick cites a 1993 paper by Torr and Li that I have not yet evaluated. There is also a paper by Giovanni Modanese that is consistent with my own model though not identical to it. Modanese uses a perturbation theory and shows that the ground state coherence of a real superconductor does contribute to the term /zpfguv in Einstein's field equation. This is excellent for my theory. However, what Modanese does not have is the key idea that there is off-mass-shell vacuum coherence from a condensate of virtual electron-positron pairs that can beat against that superconducting ground state coherence from a condensate of on-mass-shell electron pairs. The attractive virtual photon exchange between the virtual electron-positron pairs inside the physical vacuum plays the same role as the attractive virtual phonon exchange between the real electrons forming a bound state pair. The emergent collective coherence fields do not remember their micro-dynamics at the lower level and Modanese's model suggests that they interfere because both contribute to the same /zpfguv term in Einstein's field equation for quantum fields on a curved space-time background. > This, in principle, can explain the sort of thing Podkletnov has reported. Whether or not his particular experiment is any good I do not know, but it does not matter since the flying saucer data show that what I am talking about has already been implemented in nuts and bolts technology, though probably not by men who are of women born. > > PS. Modanese also does not realize that Einstein's field equations themselves can be mathematically derived from the More is different collective emergence of the physical vacuum coherence. My basic equation, in more specialized form, has also been independently derived from microscopic quantum dynamics of interacting bosons by a Russian physicist, G.E. Volovik and published in a book The Universe in a Helium Droplet. These ideas are completely absent in the official NASA BPP PV theory by Hal Puthoff & Co. > > See http://www.americanantigravity.com/podkletnov.html I am not, at this time, endorsing any of the claims or papers on this web page. > > I am suspicious of Ning Li's electro-gravitic claims in http://popularmechanics.com/science/research/1999/10/taming_gravity/print.ph t ml > > but I cannot dispute them because I have not yet carefully studied them. It sounds like a torsion field theory, i.e. influence of quantum spins of matter on the geometrodynamic field. This does happen when one locally gauges the Lorentz group in addition to the translation group. However what is suspicious in Ning Li's claim is that the strength of this spin-gravity coupling is so large. > > Nick Cook cites some really strange extraordinary claims that I am extremely skeptical of at the moment of writing this: > > gravity reduction would not diminish with distance, that the effect had no limit. Podkletnov's gravity shield went on extending upward in a 30 cm diameter column - forever. p. 106 > > Podkletnov claims a 2% effect. Podkletnov does not seem to understand warp drive (Paul Hill's acceleration field), is thinking only of flight through air? That is, > > a vehicle equipped with gravity shielding would be able to levitate, buoyed up by the heavier air below p. 106 > > This is not good enough. The real thing works in a vacuum and you do not need air. Furthermore, a real anti-gravity field from negative zero point quantum pressure will cause and anomalous blue shift. If this 2% effect is real such a blue shift should be detectable if the proper measurements are made from spectroscopy of excited atoms placed inside the alleged infinite 30 cm cylinder. Of course none of the people working on that thought of this prediction I have just made here for the first time in the context of those allegations reported by Nick Cook. > > Podkletnov's claims about reflecting gravity waves (p. 106) appear to be nutty. Ask Kip Thorne at Cal Tech to explain why. Podkletnov's remark > like UFOs, I have achieved impulse reflection shows to my mind that he does not understand the weightless warp drive idea shown in George Trimble's G-Engine quote that Nick Cook cites, in Paul Hill's acceleration field, in the Bondi-Terletskii negative matter propulsion, in the timelike geodesic Alcubierre warp drive model all culminating in the bottled dark zero point energy of negative quantum pressure that I am talking about. > > Had Podkletnov tripped over the answer...? If anyone knew, it was Dr. Hal Puthoff ... p. 107 > > Surely you must be joking Nick? I beg to differ. ;-) > === Subject: SL(n,K) perfect -> PSL(n,K) simple? Is there a relatively elementary way to show that PSL(n,K) is simple when SL(n,K) is perfect, without invoking the whole Lie-group machinery of BN pairs, Tits systems, etc.? By a relatively elementary way I mean one that's on roughly the same level as showing that SL(n,K) is (almost always) perfect, by showing that it's generated by its transvections and that these are (almost always) commutators. -- Jim Heckman === Subject: Trying to solve the parameter to the 8th I. ASPECTS OF 8th DEGREE POLYNOMIAL i. a[8]x^8+a[7]x^7+a[6]x^6+a[5]x^5+a[4]x^4+a[3]x^3+a[2]x^2+a[1]x^1+a[0]x^0=0 x = x ii. x = {(x^8)(x^7)(x^6)(x^5)(x^4)(x^3)(x^2)(x^1)}^1/36 iii. x = {(1/4)[(x^8)(x^1) + (x^7)(x^2) + (x^6)(x^3) + (x^5)(x^4)]}^1/9 Equating ii. and iii, (x^8)(x^7)(x^6)(x^5)(x^4)(x^3)(x^2)(x^1) = {(1/4)[(x^8)(x^1) + (x^7)(x^2) +(x^6)(x^3)+ (x^5)(x^4)]}^4 Giving each x^k their repectable coefficient from i. [a[1]*a[2]*a[3]*a[4]*a[5]*a[6]*a[7]*a[8] ](x^36) = {(1/4)[a[8]a[1] + a[7]a[2] + a[6]a[3]+ a[5]a[4] ]}^4 (x^36) + D x = { 1 / [ [a[1]*a[2]*a[3]*a[4]*a[5]*a[6]*a[7]*a[8] ] - {(1/4)[a[8]a[1] + a[7]a[2] + a[6]a[3]+ a[5]a[4] ]}^4 ] }^1/36 + D where D is repositioned Let C = 1 / [ [a[1]*a[2]*a[3]*a[4]*a[5]*a[6]*a[7]*a[8] ] - {(1/4)[a[8]a[1] + a[7]a[2] + a[6]a[3]+ a[5]a[4] ]}^4 ] (x - D)^36 = C The first term of the Maclaurin expansion of f(x) = (x - D)^8 - C^(8/36) = 0 is, f(0) = (-D)^8 - C^(8/36) = a[0] D = - {a[0] + C^(8/36) }^1/8 x - D = C^1/36 x = C^1/36 - (a[0] + C^(8/36) }^1/8 For further terms, f(0) = D^8 - C^(8/36) f '(0) = -8 D^7 f ''(0) = 7*8 D^6 f '''(0) = -6*7*8 D^5 f ''''(0) = 5*6*7*8 D^4 f^{5}(0) = -4*5*6*7*8 D^3 f^{6}(0) = 3*4*5*6*7*8 D^2 f^{7}(0) = -8! D f^{8}(0) = 8! a[0]/a[8] = D^8 - C^(8/36) a[1]/a[8] = -8 D^7 a[2]/a[8] = ((7*8)/2!) D^6 a[3]/a[8] = -((6*7*8)/3!) D^5 a[4]/a[8] = ((5*6*7*8)/4!) D^4 a[5]/a[8] = -((4*5*6*7*8)/5!) D^3 a[6]/a[8] = ((3*4*5*6*7*8)/6!) D^2 a[7]/a[8] = -((8!)/7!) D a[8]/a[8] = (8!)/8! D^8 = (a[0]/a[8]) + C^(8/36) D^7 = -(a[1]/a[8])(1/8) D^6 = (a[2]/a[8])(2!/(7*8)) D^5 = -(a[3]/a[8])(3!/(6*7*8)) D^4 = (a[4]/a[8])(4!/(5*6*7*8)) D^3 = -(a[5]/a[8])(5!/(4*5*6*7*8)) D^2 = (a[6]/a[8])(6!/(3*4*5*6*7*8)) D^1 = -(a[7]/a[8])(7!/8!) Either there are constraints on the coefficients a[k], or the system is underdetermined and needs to accomodate more parameters (more constants), as D was added, or the system describes 8 different values of D (36 with DeMoivre), resulting in 8 different values of x (roots). Let b[0] = { (a[0]/a[8]) + C^(8/36)}^1/8 b[1] = {-(a[1]/a[8])(1/8)}^1/7 b[2] ={ (a[2]/a[8])(2!/(7*8))}^1/6 b[3] ={-(a[3]/a[8])(3!/(6*7*8))}^1/5 b[4] = { (a[4]/a[8])(4!/(5*6*7*8))}^1/4 b[5] = {-(a[5]/a[8])(5!/(4*5*6*7*8))}^1/3 b[6] ={ (a[6]/a[8])(6!/(3*4*5*6*7*8))}^1/2 b[7] = {-(a[7]/a[8])(7!/8!)} The expression for D should include contributions from all b[k] above. Since D is conducive to any operation, so long as it is included in the final solution, suppose that D is the solution to the linear system, (b[0]^1)x[0] + (b[1]^1)x[1] + (b[2]^1)x[2] + ... + (b[7]^1)x[7] = D (b[0]^2)x[0] + (b[1]^2)x[1] + (b[2]^2)x[2] + .... = D (b[0]^3)x[0] + ... . . . . b[0]^7)x[0] + ... +(b[7]^7)x[7] = D alternatively, (b[0]^1)x[0] + (b[1]^1)x[1] + (b[2]^1)x[2] + ... + (b[7]^1)x[7] = D (b[0]^2)x[0] + (b[1]^2)x[1] + (b[2]^2)x[2] + .... = D^2 (b[0]^3)x[0] + ... . . . . . (b[0]^7)x[0] + ... +(b[7]^7)x[7] = D^7 Where x[r] are solved for using matrix algebra. x[r] are not to be confused with the x below. Once D is found, it is substituted in, x = { 1 / [ [a[1]*a[2]*a[3]*a[4]*a[5]*a[6]*a[7]*a[8] ] - {(1/4)[a[8]a[1] + a[7]a[2] + a[6]a[3]+ a[5]a[4] ]}^4 ] }^1/36 + D and solved for x, one of the roots to the 8th degree polynomial. II. ANGLE FROM ARC AND CHORD Known arc A and known chord B subtend unknown angle z on a circle. Find z in terms of A and B. Bz/(2A) = sin(z/2) sin(z/2) = (z/2) - ((z/2)^3)/3! + ((z/2)^5)/5! - ((z/2)^7)/7! + ((z/2)^9)/9! - ((z/2)^11)/11! + ... B/A = 1 - ((z/2)^2)/3! + ((z/2)^4)/5! - ((z/2)^6)/7! + ... Letting x = (z/2)^2, (1-B/A) - x/3! +(x^2)/5! - (x^3)/7! + (x^4)/9! - (x^5)/11! + (x^6)/13! - (x^7)/15! + (x^8)/17! = 0 a[8]x^8+a[8]x^7+a[6]x^6+a[5]x^5+a[4]x^4+a[3]x^3+a[2]x^2+a[1]x+a[0] a[8] = 1/17! a[7] = -1/15! a[6] = 1/13! a[5] = -1/11! a[4] = 1/9! a[3] = -1/7! a[2] = 1/5! a[1] = -1/3! a[0] = (1-B/A) C = 1 / [ [a[1]*a[2]*a[3]*a[4]*a[5]*a[6]*a[7]*a[8] ] - {(1/4)[a[8]a[1] + a[7]a[2] + a[6]a[3]+ a[5]a[4] ]}^4 ] [ 1/(3!*5!*7!*9!*11!*13!*15*17!) ] = 4.09025223E-47 [(1/17!)(-1/3!)+(-1/15!)(1/5!)+ (1/13!)(-1/7!)+ (-1/11!)(1/9!)] =1.077412897E-13 C = 4.09025218E-47 Using the first term only of the Maclaurin expansion, x = C^1/36 - {1-B/A + C^(8/36) }^1/8 x = (4.09025218E-47)^1/36 - {1-B/A + (4.09025218E-47)^(8/36)}^1/8 Suppose B=2^1/2 and A=pi/2 Solving for z, 1-B/A = 1-(2*2^1/2)/pi = .0997 x = (4.09025218E-47)^1/36 - {.0997 + (4.09025218E-47)^(8/36)}^1/8 .05145 .74961 x = -.69815 | x | = .69815 z = 1.67111 = 0.5319 pi the answer should be 0.5 pi. Roughly, angle z subtends arc A and chord B according to the relation, z = 2{ | C^(1/36) - {1-B/A + C^(2/9)}^1/8 | }^1/2 Where C = 1 / [ 1/(3!*5!*7!*9!*11!*13!*15*17!) - {(1/4)[(1/17!)(-1/3!)+(-1/15!)(1/5!)+(1/13!)(-1/7!)+(-1/11!)(1/9!)]}^4 ] Jon Giffen === Subject: Re: P vs NP: my proof of P != NP. oo ____|mn / /_/ / _ / K-9/ /_/ - sci.math & comp.theory added - /____/_____ -------------- > Hi All. > My name is Mikhail N. Kupchik, I am 3rd year undergraduate of > department of applied mathematics - university of KPI, Kiev, Ukraine. > I've written a proof for the subject problem - 21 pages paper - and > I'd like to know public opinion of it. The work was written and > http://users.i.com.ua/~zkup/pvsnp_en.pdf . (But please don't download > the file until you're sure you'll read all the details. First read the > sketch below.) > The schema of the proof is the following. > 1. I am building a problem, called AEASAT - a decision problem which > decides is given formula of the form > forall x_1 forall x_2 ... forall x_n > exists y_1 exists y_2 ... exists y_m > forall z_1 forall z_2 ... forall z_k > F[ x_1 ... x_n, y_1 ... y_m, z_1 ... z_k ] > is true or false. > Clearly if P=NP then AEASAT is in P. And degree of time-complexity > must be fixed, i.e. independent of n,m,k and length of F. > 2. Then I prove that _every_ problem from EXP is P-reducible to > AEASAT. > I am building formal description of Turing Machine work, add initial > conditions, the result (I consider decision problems only) and then > trying to move all quantifiers ahead. (I am building a sequence of > equivalent expressions). The length of core expression during such > 'move' must grow no more than a fixed-degree polynomial, that is > prooved too. (Core means quantifier-free inner part). Most part of the > paper consists of a techical details of this step. > 3. (1)&(2) implies P = NP -> EXP = P. > Because we know EXP != P (by Hartmanis,Stearns), then P != NP. > Unfortunately last step was too hard for someone who reviewed it > before (I don't want to name them here). > There is a russian version also: > http://users.i.com.ua/~zkup/pvsnp_ru.pdf === Subject: Scientifically Historical Blindness Comments: This message probably did not originate from the above address. It was automatically remailed by one or more anonymous mail services. You should NEVER trust ANY address on Usenet ANYWAYS: use PGP !!! Get information about complaints from the URL below unreachable: It is because it has been flooded by , please contact -----BEGIN PGP SIGNED MESSAGE----- Until the days of Copernicus, when perpetual thick clouds had enveloped the entire Earth since the dawn of man 10s or maybe 100s of millions of sidereal Earth-years before the sixteenth century anno Domini, no human being could ever view the night sky for the dense cloud-cover obscuring the caelestial heaven. As a result, all textbooks properly report that Copernicus is the father of astronomy, the first astronomer in all of human history, the first man to see beyond the cloud cover, because until Copernicus was born JD 2259121.14830, 18E36:38 53N01:18, no man had ever been able to see the clear & starry night sky. At least, that's what EVERY accredited university is teaching the world's children! Every classroom, from elementary school through high school, every institution of higher learning has taught that Copernicus is the first man in history to see the clear night sky beyond the clouds. Is this true? Ask yourself, is Copernicus the first man EVER to see the stars and planets? Since it's irrefutably proven that anyone who can see the sky on occasionally clear nights can easily estimate the sidereal and synodic orbital periodicities of the planets, these facts prove that before Copernicus, no one could ever see the night sky--or if they could, then the Copernican Revolution is dead. The evidence is crystal clear. Clouds used to cover the Earth at all times before Copernicus came along...is it that simple? *Min's $0.00 (zero cents, was cash $10,000 U.S.) Reward For Orthodox Astronomers: *Think* About It, Daniel Joseph Min ________________________________________________ LINKS TO FINE ON-LINE BOOKS BY DANIEL JOSEPH MIN ________________________________________________ *Min's Interlinear Apocalypse & Concise Commentary (Min's fifth book): *Min's Planetary Awareness Technique (Min's fourth book): *Min's Light & Sound (Min's third book): *Min's Compleat Tarot Manual (Min's second book): *Min's Historical Calendar Of Jesus (Min's first book): *Min's Astro-Charts & Tarot Pages: *Min's Beginner's Reference & FAQs: *Min's Nostradamus Interlinear Translations: *Min's Horoscopes & Souljourns of World-Famous Persons: *Min's Horoscopes & Souljourns of Unknowns & Anonymous: *Min's Official PGP Public Key on the MIT server: *Min's Newsgroup-Archived Home Page On The World Wide Web: -----BEGIN PGP SIGNATURE----- iQA/AwUBQHI6bJljD7YrHM/nEQKreACfXhgOo+p3cVEnbL4scgeBZDHZD5wAniai eHuchfdKqmLaSe0vK8wfvYRQ =o0X2 -----END PGP SIGNATURE----- === Subject: Re: PROOF that numbers are countable [snip lots] >> How about going back to my previous post, correcting my typo, and trying >> again. >> You (like other Phillites) are sorely confused about the difference between >> Infinite items and Infinite SETS of finite items (e.g. natural numbers, >> sequence of 3's). My concrete example was an attempt to show you that >> difference in your own terms, but there are none so blind.... >have you wondered why you're the only person left arguing Cantors theorem? No, actually. I expect that most posters have given up on you as a lost cause. To be frank, so have I; but I still find some of your contortions entertaining. BTW, you seem to have lost the thread: what I've been arguing here is that Phui is false, i.e. that there is a BIG difference between P is true for all finite sequences, no matter how large and P is true for infinite sequences. >>S is some finite or infinite set of reals >>Ai e S, i<(1/3) <-> Ej, iWhat about this then? >S is some finite or infinite set of reals >[Ai e S, i<(1/3)] <-> [Ej, Ai e S, i [snip for brevity] >> You (like other Phillites) are sorely confused about the difference between >> Infinite items and Infinite SETS of finite items (e.g. natural numbers, >> sequence of 3's). My concrete example was an attempt to show you that >> difference in your own terms, but there are none so blind.... > have you wondered why you're the only person left arguing Cantors theorem? > For some version of only. :-) >S is some finite or infinite set of reals >>Ai e S, i<(1/3) <-> Ej, i [Ai e S, i<(1/3)] <-> [Ej, Ai e S, i That might work. Of course j will have to be restricted > to the rationals (a countable set) for this to be of any > interest at all. S = { 0.3 0.33 0.333 0.3333 ... } Is [Ej, Ai e S, iSuppose I take the ZFC definition of infinite sets: >An infinite set I is a set for which there is a proper subset T and an >invertible map I->T . >(by the way, can this be simplified to a one-to-one map I->T?) By invertible, do you mean a one-to-one correspondence? I guess from the sequel the answer is yes. The answer to your parenthetical question is yes: The range of the map is a proper subset of I. Another way to express your definition is to say that there is a map from I to I that is one-to-one but not onto. This is the definition of Dedekind infinite, which is equivalent to infinite in ZFC. Here, I use infinite to mean not finite, where finite means bijective with a natural number (i.e, an element of omega). >In any case, I am writing a document about cardinality for the benefit >of some people (and practicing proving stuff along the way). I had >defined the following: >|A| <= |B| says that there is a one-to-one function from A into B >|A| >= |B| says there is an onto function from A into B >|A| = |B| says there is an invertible function A->B . >I was able to prove some things, including: >1) |A| <= |B| is equivalent to |B| >= |A| . >2) <= is a partial order on cardinalities (in particular, the >Cantor-Schroeder-Bernstein theorem). >Also: >A subset B implies |A| <= |B| >A is infinite and |A| <= |B| implies B is infinite >finite ^ finite = finite >finite - finite = finite >However, I am now trying to prove the following statements; they all >seem to be interrelated but I am not sure where to start. >a) finite u finite = finite >b) infinite - finite = infinite >c) every infinite set contains a countable subset >d) all finite sets have a cardinality in N >that last statement might read, the set of cardinal numbers of finite >sets (i.e. not infinite sets) is isomorphic to N. >Do I need the definition of N={0, 1, ...} for these things? Maybe I >can't prove them without defining N to be the set of all cardinal >numbers of finite sets. >Basically, I am curious how to prove a-d and what I would need for >them. How do I show that the cardinality of the integers (omega?) is >the next highest after finite cardinalities and that all the finite >sets have a number of elements that is in N? You should not need a definition of N to prove (a) and (b). Without defining N, how can you even speak of countable in (c) or N in (d)? After all, the Axiom of Infinity is in ZF - why not use it? I think it would help you to take a look at the development in Halmos, Naive Set Theory. After that, you can look at a more advanced book such as Drake or Kunen. The first chapter of Munkres' Topology text also has everything you need, I think. Lately, I have been thinking about infinite vs. Dedekind infinite *without* the Axiom of Choice. See, for example, These exercises might help you in your endeavor as well. Here is my latest pondering: Say X has property (*) iff there exists a onto map from X to N. Can one show in ZF - if a set is infinite, it has property (*)? - if a set has property (*), it is Dedekind infinite? I know that one cannot show that both implications are true without choice, for infinite does not imply Dedekind infinite in ZF. -- Stephen J. Herschkorn herschko@rutcor.rutgers.edu === Subject: Re: Theorems about Infinite sets <4070EEF2.5080705@rutcor.rutgers.edu Lately, I have been thinking about infinite vs. Dedekind infinite > *without* the Axiom of Choice. See, for example, > These exercises might help you in your endeavor as well. Here is my > latest pondering: Say X has property (*) iff there exists a onto map > from X to N. Can one show in ZF > - if a set is infinite, it has property (*)? > - if a set has property (*), it is Dedekind infinite? > I know that one cannot show that both implications are true without > choice, for infinite does not imply Dedekind infinite in ZF. My guess is that neither implication is provable in ZF, but that stuff is really out of my line. But here are some things that are easy to prove in ZF. P(X) is the power set of X. If X is infinite, then P(X) has property (*). If X has property (*), then P(X) is Dedekind infinite. === Subject: Re: Theorems about Infinite sets > Suppose I take the ZFC definition of infinite sets: > > An infinite set I is a set for which there is a proper subset T and an > invertible map I->T . > > (by the way, can this be simplified to a one-to-one map I->T?) > > In any case, I am writing a document about cardinality for the benefit > of some people (and practicing proving stuff along the way). I had > defined the following: > > |A| <= |B| says that there is a one-to-one function from A into B > |A| >= |B| says there is an onto function from A into B > |A| = |B| says there is an invertible function A->B . > > I was able to prove some things, including: > > 1) |A| <= |B| is equivalent to |B| >= |A| . > 2) <= is a partial order on cardinalities (in particular, the > Cantor-Schroeder-Bernstein theorem). > > Also: > > A subset B implies |A| <= |B| > A is infinite and |A| <= |B| implies B is infinite > > finite ^ finite = finite > finite - finite = finite > > However, I am now trying to prove the following statements; they all > seem to be interrelated but I am not sure where to start. > > a) finite u finite = finite > b) infinite - finite = infinite > c) every infinite set contains a countable subset > d) all finite sets have a cardinality in N > > that last statement might read, the set of cardinal numbers of finite > sets (i.e. not infinite sets) is isomorphic to N. > > Do I need the definition of N={0, 1, ...} for these things? Maybe I > can't prove them without defining N to be the set of all cardinal > numbers of finite sets. > Read Jech's discussion and series of exercises in Set Theory. You define N to be the set of ordinals smaller than the smallest limit ordinal. Then using the axiom of choice you can prove your desired results. > Basically, I am curious how to prove a-d and what I would need for > them. How do I show that the cardinality of the integers (omega?) is > the next highest after finite cardinalities and that all the finite > sets have a number of elements that is in N? > > Anyway, stuff like that. > > -Greg === Subject: Re: Deep Thoughts # 3: Researchers Haven't Formalized the Basics (Proposed Solution) it suffers the same problem with every other notation system, that will > like many others also cause it to not be picked up. its completely > ambiguous. Example? > reading the document should be explicit without a legend. > > rec(f), pro(n, f).. > > good points on formalism, we forget godels proof when we dictate formalism. > almost none of what we write is formal, we just base arguments over formal > constructs. > > Herc Defining YES(a,b) to mean Turing Machine a accepts input b., how would you formalize the fact that the set of inputs accepted by a Turing Machine is recursively enumerable? The ambiguity is in the notation you and David Ullrich are proposing. Mathematicians are used to saying things about functions (e.g. predicates), not about individual components. But that is what is needed. Charlie Volkstorf Cambridge, MA PS It's YES(I,x). Then consider formalizing Turing '37. See [http://www.arxiv.org/html/cs.lo/0003071]. === Subject: Re: Deep Thoughts # 3: Researchers Haven't Formalized the Basics (Proposed Solution) > >Deep Thoughts # 3: Researchers Haven't Formalized the Basics (Proposed >Solution) 1. The reason that research into Logic and Theoretical Computer >Science is so deficient > > [deep thoughts on explanation of definiciencies in everyone > else's thinking snipped, so we can get to the content] > Let f be any one-place function, P be any set and N be any program. >Then we can define: f(I) f is recursive >P(I) P is recursive >P(x) P is recursively enumerable. >N # f(I) Program N computes function f > > Golly, that _is_ deep. We can invent symbols for these concepts. > > Wow. > > (Of course something like I(f) would make more sense than > f(I) here, How's that? Simple example: Let YES(a,b) mean Turing Machine a halts yes on input b. and consider the fact that for any Turing Machine the set of inputs on which it halts yes is recursively enumerable. This is formalized as YES(I,x). How would you represent this if the function (predicate) YES is the argument of I? The problem with your notation is that the individual components represent the recursive or recursively enumerable set, not the entire function. You need to make the components I or X, not the function. > but never mind - No, tell me. Why would I(f) make more sense? How would you represent the above simple example? > that's the sort of quibble that > small minds come up with when they're stunned by awesome > breakthroughs like this.) > > ************************ > > David C. Ullrich So you agree with all of the results in my papers cited e.g. [http://www.arxiv.org/html/cs.lo/0003071]? (I assume that you actually read a paper before you bash it, don't you?) Charlie Volkstorf Cambridge, MA === Subject: Another combinatoric problem.... Consider a set of n objects that are different. Suppose we want to divide this set of n objects into k subsets, and the ordering of the subsets matter, i.e., (a, b), (c) is different from (c), (a, b) If empty set is allowed, the number of ways is k^n, because for each of the n objects, we can choose to put it into one of the k sets. This is a bit different from Stirling Number of the Second Kind, becuase empty set is allowed, and then the ordering of the set matters now. Now, suppose we have the restriction that the largest subset must have size less than m, where m < n. What is the number of ways to divide this n objects?